How do you find the vertex and the intercepts for $f (x) = - x^{2} + x - 5$ $f(x) = -x^2 + x - 5$ ?

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How do you find the vertex and the intercepts for $f (x) = - x^{2} + x - 5$ $f(x) = -x^2 + x - 5$ ?

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Answer 1 · 2017-05-17T12:46:15

Vertex: $= (\frac{1}{2}, 5 \frac{3}{4})$ $=(1/2, 5 3/4)$
$y$ $y$ -intercept: $x = 0$ $x=0$ , $y = - 5$ $y=-5$
$x$ $x$ -Intercept: none

Explanation:

This equation is in standard form
#y=ax^2+bx+c

Where $a = - 1$ $a=-1$ , $b = 1$ $b=1$ , and $c = - 5$ $c=-5$

Vertex: $(- \frac{b}{2} a, f (- \frac{b}{2} a))$ $(-b/2a, f (-b/2a))$
$= (\frac{1}{2}, 5 \frac{3}{4})$ $=(1/2, 5 3/4)$

Intercepts:
When $x = 0$ $x=0$ , $y = - 5$ $y=-5$
When $y = 0$ $y=0$ , you have to use the quadratic equation. But the number under the square root is negative, so there are no $x$ $x$ -Intercepts.

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How do you find the vertex and the intercepts for $f (x) = - x^{2} + x - 5$ $f(x) = -x^2 + x - 5$ ?

1 Answer

Explanation:

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How do you find the vertex and the intercepts for f(x) = -x^2 + x - 5f(x)=−x2+x−5f(x) = -x^2 + x - 5?

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Explanation:

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How do you find the vertex and the intercepts for $f (x) = - x^{2} + x - 5$ $f(x) = -x^2 + x - 5$ ?